To measure the internal resistance of a battery, potentiometer is used. For R = 10 Ω, the balance point is observed at l = 500 cm and for R = 1 Ω the balance point is observed at l = 400 cm. The internal resistance of the battery is approximately: (a) 0.1 Ω (b) 0.3 Ω (c) 0.2 Ω (d) 0.4 Ω

Answer: (b). l = l₀R/R+r: 500 = l₀(10)/(10+r), 400 = l₀1/1+r. Dividing: 5/4 = (10(1+r))/(10+r) ⇒ 50+5r = 40+40r r = 10/35 ≈ 0.29 Ω So the answer is (b) 0.3 Ω.

To measure the internal resistance of a battery, potentiometer is… | JEE Main 2024 (4 Apr, Shift 1)

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Potentiometer

To measure the internal resistance of a battery, potentiometer is used. For $R = 10\ \Omega$ , the balance point is observed at $l = 500\text{ cm}$ and for $\mathrm{R} = 1\ \Omega$ the balance point is observed at $l = 400\text{ cm}$ . The internal resistance of the battery is approximately:

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Answer: (b)

$l = l_0\dfrac{R}{R+r}$ : $500 = l_0\dfrac{10}{10+r}$ , $400 = l_0\dfrac{1}{1+r}$ .
Dividing: $\dfrac{5}{4} = \dfrac{10(1+r)}{10+r} \Rightarrow 50+5r = 40+40r$
$r = \dfrac{10}{35} \approx 0.29\ \Omega$
So the answer is (b) $0.3\ \Omega$ .

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